Find the equation of the normal to the curve | vedclass

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(A) Differentiating $x^{2}=4y$ with respect to $x$,we get $\frac{dy}{dx}=\frac{x}{2}$.Let $(h, k)$ be the coordinates of the point of contact of the n

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$x+y=3$

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(A) Differentiating $x^{2}=4y$ with respect to $x$,we get $\frac{dy}{dx}=\frac{x}{2}$.Let $(h, k)$ be the coordinates of the point of contact of the normal to the curve $x^{2}=4y$. The slope of the tangent at $(h, k)$ is $\frac{h}{2}$.Thus,the slope of the normal at $(h, k)$ is $m = -\frac{2}{h}$.The equation of the normal at $(h, k)$ is $y-k = -\frac{2}{h}(x-h)$.Since the normal passes through $(1, 2)$,we have $2-k = -\frac{2}{h}(1-h)$,which simplifies to $k = 2 + \frac{2}{h}(1-h) = 2 + \frac{2}{h} - 2 = \frac{2}{h}$.Since $(h, k)$ lies on the curve $x^{2}=4y$,we have $h^{2}=4k$. Substituting $k = \frac{2}{h}$,we get $h^{2} = 4(\frac{2}{h}) = \frac{8}{h}$,so $h^{3}=8$,which gives $h=2$.Then $k = \frac{2}{2} = 1$.The equation of the normal at $(2, 1)$ is $y-1 = -\frac{2}{2}(x-2)$,which simplifies to $y-1 = -(x-2)$,or $x+y=3$.

Find the equation of the normal to the curve $x^{2}=4y$ which passes through the point $(1,2)$.

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